344. 反转字符串 - 力扣（LeetCode）

方法一

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        left = 0
        right  = len(s)-1
        while left < right:
            temp = s[left]
            s[left] = s[right]
            s[right] = temp

            left += 1
            right -= 1

        return s

方法二

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        n = len(s)
        for i in range(n//2):
            s[i], s[n-i-1] = s[n-i-1], s[i]

方法三

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        s[:] = s[::-1]

• 刷题的时候不要图方便，理解库函数是怎么运作再用

541. 反转字符串 II - 力扣（LeetCode）

class Solution(object):
    def reverseStr(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: str
        """
        def reverseString(s):
            """
            Do not return anything, modify s in-place instead.
            """
            left = 0
            right  = len(s)-1
            while left < right:
                temp = s[left]
                s[left] = s[right]
                s[right] = temp #可以不用中间值，可以写在一行

                left += 1
                right -= 1

            return s

        s = list(s)
        n = len(s)
        for i in range(0, n, 2*k): 
            if n-i >= k: 
                s[i:i+k] = reverseString(s[i:i+k]) #可以只写这一步，因为切片就算取不到i+k也会返回到最后一位
            else: 
                s[i:n] = reverseString(s[i:n])

        return  ''.join(s)

54.替换数字 题目页面 (kamacoder.com)

方法一，用字典记录数字，然后遍历字符串，在字典里就用number替换

s = input()
dict = {}
for x in range(10):
    dict[str(x)] = 0

s = list(s)
for i in range(len(s)):
    if s[i] in dict:
        s[i] = 'number'

s = ''.join(s)

print(s)

• 这题要回过头再看一遍

151. 反转字符串中的单词 - 力扣（LeetCode）

• python str不可变，空间复杂度不能是O(n)
• 暂时用最方便的做法

class Solution(object):
    def reverseWords(self, s):
        """
        :type s: str
        :rtype: str
        """
        s = s.split()

        s = ' '.join([word for word in s[::-1]])

        return s

55 右转字符串 题目页面 (kamacoder.com)

解题思路

和上一题一样，看作k个单词，先反转整个字符串，然后反转单词

k =int(input()) #这里要换成int型，不然会报错
s =input()

s = list(s)
s = s[::-1]
s[0:k] = s[0:k][::-1]
s[k:len(s)] = s[k:len(s)][::-1]
s = ''.join(s)
print(s)